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Let $\Omega \subset R^n$ be an open and bounded set with smooth boundary and $f \in C^2(\Omega)\cap C(\overline{\Omega})$. Let $ a \geq 0$ be a constant.

Consider the following problem:

$$ \left\{ \begin{array}{ccccccc} -\Delta u +au = f , \ in \ \Omega \\ \ u=0 , \ in \ \partial\Omega \\ \end{array} \right. $$

Can someone give me a reference for the proof of the existence of classical solution for this problem?

In this text http://www3.nd.edu/~qhan/PDE.pdf, it discusses maximum principles about the problem that I said above. but I haven't found a book with the proof of the existence.

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  • $\begingroup$ Is there nothing about $f$ on the boundary? $\endgroup$ – Tomás Oct 12 '13 at 16:11
  • $\begingroup$ Yes. I forget : $f \in C(\overline{\Omega})$ . sorry $\endgroup$ – math student Oct 12 '13 at 18:05
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$f\in C(\overline{\Omega})$ implies that $f\in L^2(\Omega)$ which implies that there exist unique weak solution $u\in H_0^1(\Omega)\cap H^2(\Omega)$.

Moreover, $f\in C(\overline{\Omega})$ implies that $f\in L^p(\Omega)$ for all $p$, which implies that $u\in W_0^{1,p}(\Omega)\cap W^{2,p}(\Omega)$ for all $p$ (see Theorem 9.32 of Brezis book).

We conclude from the last paragraph that $u\in C^{1,\alpha}(\overline{\Omega})$ for some $\alpha\in (0,1)$. To conclude note that $f\in C^2(\Omega)$ implies interior regularity (see section 9.6 of Brezis book) and therefore $u\in C^2(\Omega)$.

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  • $\begingroup$ for the implication $f \in C^{2}(\Omega)$ implies $u \in C^2 (\Omega)$ wich theorem do you used of the section 9.6 ? $\endgroup$ – math student Oct 12 '13 at 19:50
  • $\begingroup$ I suggest you to read all of the section 9.6. After it, I think you gonna find it. $\endgroup$ – Tomás Oct 12 '13 at 20:08
  • $\begingroup$ Eu dei uma olhada por cima na seção. aí fiquei com medo de a justificativa para a implicacao que falei acima ser algo muito escondido na seção. Olharei a seçao com mais calma . obrigado pela ajuda ! $\endgroup$ – math student Oct 12 '13 at 20:13
  • $\begingroup$ De nada, pode procurar la que vc acha, tá no meio de uma demonstração. $\endgroup$ – Tomás Oct 12 '13 at 20:13
  • $\begingroup$ Em relação a justificativa: por acaso a justificativa é a hipoeliticidade mencionada na observação 25? $\endgroup$ – math student Oct 12 '13 at 21:14

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